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  7. Coxeter Groups are Biautomatic: Lecture 2 by Piotr Przytycki & Damian Osajda

Coxeter Groups are Biautomatic: Lecture 2 by Piotr Przytycki & Damian Osajda

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Learn about the concept of Coxeter groups being biautomatic in this informative lecture by Piotr Przytycki and Damian Osajda. Discover the criteria for a group to be considered biautomatic and the properties it entails, such as having a generating set and regular language. Dive into the details of how every group element has a representative in the language and the fascinating fellow traveler property.
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Video Summary & Chapters

Video Transcript

0:11
first talk of today is uh P's second
0:14
talk on Cox to groups being biomatic
0:17
thank
0:18
you so first a small recap from last
0:22
time a group is by automatic if there is
0:24
a generating set s and the regular
0:27
language over s such that the map from
0:31
the language to the group is subjective
0:33
every word has a represent every group
0:35
element has a representative in this
0:38
language and fellow traveler property
0:40
holds so for every pair of words in the
0:44
language that are equal in the group or
0:47
the fair by a generator on the right
0:51
these words feral travel and second part
0:54
if they defer on the left in the group
0:56
then the words W and s w Prime below
0:59
travel
1:01
and we Pro the corollary last time which
1:03
is true even for automatic groups so
1:05
just for groups satisfying part one that
1:10
one implies
1:13
that any word any group
1:16
element in
1:18
the uh automatic group represented by a
1:21
word that might not necessarily be in
1:23
the language can be still represented
1:25
efficiently can be represented by a word
1:27
in the language of length at most some
1:30
constant times the length of the
1:31
original word plus and
1:33
zero and today we are supposed to prove
1:36
the theorem that every automatic group
1:38
so satisfying just part
1:41
one is finally presented we assume it's
1:44
finally generated but for the moment we
1:45
don't know it's Finly presented and
1:49
moreover it satisfies quadratic
1:51
isoparametric inequality which means
1:53
that there is a constant C such that for
1:55
any word representing the trial element
1:58
in the group this word can be
2:00
represented efficiently as a product of
2:03
conjugates of
2:04
relators in the free group namely the
2:07
number of these conjugates it say most
2:10
quadratic in the length of the
2:13
word any questions before we start the
2:15
proof okay so let's do the
2:23
proof so we have a

Video Summary & Chapters

Video Transcript

0:11
first talk of today is uh P's second
0:14
talk on Cox to groups being biomatic
0:17
thank
0:18
you so first a small recap from last
0:22
time a group is by automatic if there is
0:24
a generating set s and the regular
0:27
language over s such that the map from
0:31
the language to the group is subjective
0:33
every word has a represent every group
0:35
element has a representative in this
0:38
language and fellow traveler property
0:40
holds so for every pair of words in the
0:44
language that are equal in the group or
0:47
the fair by a generator on the right
0:51
these words feral travel and second part
0:54
if they defer on the left in the group
0:56
then the words W and s w Prime below
0:59
travel
1:01
and we Pro the corollary last time which
1:03
is true even for automatic groups so
1:05
just for groups satisfying part one that
1:10
one implies
1:13
that any word any group
1:16
element in
1:18
the uh automatic group represented by a
1:21
word that might not necessarily be in
1:23
the language can be still represented
1:25
efficiently can be represented by a word
1:27
in the language of length at most some
1:30
constant times the length of the
1:31
original word plus and
1:33
zero and today we are supposed to prove
1:36
the theorem that every automatic group
1:38
so satisfying just part
1:41
one is finally presented we assume it's
1:44
finally generated but for the moment we
1:45
don't know it's Finly presented and
1:49
moreover it satisfies quadratic
1:51
isoparametric inequality which means
1:53
that there is a constant C such that for
1:55
any word representing the trial element
1:58
in the group this word can be
2:00
represented efficiently as a product of
2:03
conjugates of
2:04
relators in the free group namely the
2:07
number of these conjugates it say most
2:10
quadratic in the length of the
2:13
word any questions before we start the
2:15
proof okay so let's do the
2:23
proof so we have a
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